ece606: solid state devices lecture 14 …ee606/downloads/ece606_f...klimeck –ece606 fall 2012...

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

ECE606: Solid State DevicesLecture 14

Electrostatics of p-n junctions

Gerhard [email protected]


Outline

2

1) Introduction to p-n junctions

2) Drawing band-diagrams

3) Accurate solution in equilibrium

4) Band-diagram with applied bias

Ref. Semiconductor Device Fundamentals, Chapter 5


What is a Diode good for ….

3

solar cells GaAs lasers

GaN lasersAvalanche Photodiode

Organic LED

Image.google.com


p-n Junction Devices …

4

DN

AN

Schematic of a p-n Diode

Contact

Contact

P-doped pocket

N-dopedsubstrate

N P

Symbol

Point-contact diode


Topic Map

5

Equilibrium DC Small signal

Large Signal

Circuits

Diode

Schottky

BJT/HBT

MOS

Topic Map (Today : Diode in Equilibrium)

Diode in Equilibrium.(No external voltage applied)


Drawing Band Diagram in Equilibrium…

6

1∂ −= ∇ • − +∂

JP P P

pr g

t q

( )+ −∇ • = − + −D AD q p n N N

P P Pqp E qD pµ= − ∇J

1N N N

nr g

t q

∂ = ∇ • − +∂

J

J µ= + ∇N N Nqn E qD n

Equilibrium(Start here)

Non-Equilibrium (refine later)DC dn/dt=0

Small signal dn/dt ~ jωt x n Transient --- full solution

Previously constant in homogeneous semiconductors. But for pn diode: f(x) !!

In equilibrium J=0 (no current flow).But, Electric fields or diffusion might still be present. � Detailed balance


P and N doped Material Side by Side …

7

N-doped

Conduction Band

Valence Band

Ec

EV

EF

k-k

E

P-doped

Conduction Band

Valence Band

EF

For P-doped EF close to EV

Vacuum level (determined by material)


Forming a Junction

DN AN

Space charge (fixed)Mobile carriers

Donor-side (N-side)Squares are fixed donor atoms.Every donor atom has given away one electron (blue circle)

Acceptor-side (P-side)Squares are fixed acceptor atoms.Every acceptor atom has captured one electron (from the valence band). Every acceptor atom leaves behind one hole in the valence band. (red circle)


Forming a Junction

DN AN


=D

n N

Junctionx

=A

p N2

0 = Ain n NBefore joining the p and the n side

Valid only in equilibrium


Forming a Junction

10

=D

n N

Junction

Actual Carrier Concentrations

Bulk Region Bulk RegionDepleted Region

n-side p-side

n

np

p

ln(n),ln(p)

x

x

=A

p N2

0 = Ain n N


Before joining the p and the n side

After joining the p and the n side

Depletion approximation

Depletion approximation

DN AN

Diffusion E-field


Formation of a Junction

11

DN AN+

ln(ND)

ln(n)

Q=p-n+ND-NA

Bulk Region

qND

-qNA

Bulk Region

xn

Depleted Region

xp

ln(NA)ln(p)

Net charge on N-side: qND (subtract green from blue line)Space charge region: fixed donor atomsNeglect electrons in p-side Xn: depletion region in N-side

Net charge on P-side: -qNA(subtract brown from red line)Space charge region: fixed acceptor atomsXp: depletion region in P-side

red and blue are of equal size � charge balance

Depletion Region means depleted of “mobile” charges


Sketch of Electrostatics

12

position

Potential

position

E-field

p-n+ND-NAqND

Depleted Region

position

Bulk Region

-qNA

Bulk Regionxp

max ε ε= =

os snp

oA Dx N

qE

Kx

KN

q

Vbi

xn

Integrate

Integrate


Sketch of Electrostatics

13

position

Potential

position

Band Diagram

xnxp

Invert to go from potential to energy scale

In equilibrium Fermi-level must be flat


Outline

14

1) Introduction to p-n junctions


3) Analytical solution in equilibrium



Short-cut to Band-diagram

15

DN AN

… is equivalent to solving the Poisson equation

Vacuum level

EC

EV

EF

χ2

χ1

Neutral NeutralSpace Charge

Drawing Recipe1) Start with EF2) Ec/Ev in bulk n-side3) Ec/Ev in bulk p-side4) Vacuum level in N5) Vacuum level in P6) Join vacuum levels7) “Transfer” vacuum

slopes to join Ec/Ev


Built-in Potential: boundary conditions @infinity

16

1 1 2 2 2χ χ+ + =∆ ∆+ −gb ,i EqV

2 1 12 2 χχ∆ ∆= − − + −g ,bi EqV

qVbi

∆1

∆2

χ1

χ2

( )2

11

2 2 χχ

= + + + −

B BA

g ,V , C ,

Dk T ln kN

NT ln

N

NE

( )2

2 1

12 χχ−= + −g , BB E

DT

C

A/ k

V , ,

k T lnN N e

NN

Eg,2

Always true in equilibrium

Built-in potential Vbi unknown!

delta1,2 determined via doping concentrations

X1,2 material parameters


Interface Boundary Conditions

17

position

E = (D/kεo)

xn xp

position

D

xn xp

01 0 21 2(0 ) (0 )ε ε− −= = =EKD E DK

2

1

(0 ) (0 )− +=K

E EK

Displacement is continuous across the interface, field need not be ..

Homo - Junction

Hetero - Junction

Field not continuous across junction


Built-in voltage for Homo-junctions

18

DN AN


Vacuum level

EC

EV

EF

χ1

χ1

( )2 2 1

2 1

χ χ−= + −g , Bbi B E / k T

V , C

A D

,

qV k T lnN N e

N N2−= =

g B

A AB BE / k T

C

D D

iV

k T ln k T lnnN N e

NN NN

qVbiDrawing Recipe1) Start with EF2) Ec/Ev in bulk n-side3) Ec/Ev in bulk p-side4) Vacuum level in N5) Vacuum level in P6) Join vacuum levels7) “Transfer” vacuum

slopes to join Ec/Ev

Zero for homo-junctions


Analytical Solution of Poisson Equation

19

( )2

0 2S D A

d VK q p n N N

dx+ −= − − + −ε

Q= p-n+ND-NA

qND

Depleted Region

position-qNA

xn xp


Analytical Solution for Homojunctions

20

(Charge)p-n+ND-NA

qND -qNA

xn xp

position

E-field

position

position

Potential

( ) ( )22

0 0

0 0

2 2

2 2ε ε

− +

= +

= +

n p

AD

bi

pn

s s

E x E xq

qN xxqN

k k

V

( )

( )0

0

0

0

ε

ε

−

+

=

=

⇒ =

D

s

A

s

D A

n

n

p

p

qNE

k

qN

N x

k

xE

N x

x

Vbi

Small !

Integrate

Integrate

E-field

PotentialIntegrate


Depletion Regions in Homojunctions

21

22

0 02 2ε ε= +bi

A

s

p

s

nDqNq

qk

xN

k

xV

=D pn AxN N x( )

02 ε=+

s A

D An

Dbi

k Nx

N NV

q N

( )02 ε=

+s D

A Ap

Dbi

k Nx

N NV

q N

xn xp

Small Project: Solve the same problem for a hetero-junction

DN AN


Solve for xn, xp


Complete Analytical Solution

22

N P

-xp 0 xn

x

)( AD NNq −=ρ0=ρ0=ρ

qND

-qNA

ρ

xn

-xp

E

-xp xn0

x x x

V

-xp xn

..........

................

............. ,

0

0

0

ε

ε

−− ≤ ≤

= ≤ ≤ ≤ − ≥

A

os

D

os

qNp

K

qNn

K

p n

x xd

x xdx

x x x x

E

( )

00

εε−

′ ′= −∫ ∫ p

x xA

xS

qNd dx

K

E

..........

0

( ) ( ) 0ε

= − + − ≤ ≤ppS

Aqx

Nx x x x

KE

0

( )0

εε

ε′ ′=∫ ∫

nxD

x xS

qd

Ndx

K

..........

0

( ) ( ) 0ε

= − − ≤ ≤Dnn

S

qx x x

Nx x

KE

Integrate Integrate

If you need to calculate electric field at specific points…


Outline

23

1) Introduction to p-n junction transistors


3) Analytical solution in equilibrium



Topic Map

24

Equilibrium

DC Small signal

Large Signal

Circuits

Diode

Schottky

BJT/HBT

MOS

Diode in Non-Equilibrium(External DC voltage applied)


Applying Bias to p-n Junction

25

VA

2

4

5

1

3

6,7

ln(I) 1. Diffusion limited

2. Ambipolar transport

3. High injection

4. R-G in depletion

5. Breakdown

6. Trap-assisted R-G

7. Esaki Tunneling

Forward BiasReverse Bias

IV characteristics of a DiodeTo be discussed in detail


Forward and Reverse Bias

26

DN

AN

DN

AN

Forward Bias

Reverse Bias

Current flows easily

Current does not flow easily


Band Diagram with Applied Bias…

27

1P P P

pr g

t q

∂ = − ∇ • − +∂

J

( )D AD q p n N N+ −∇ • = − + −


1N N N

nr g

t q

∂ = ∇ • − +∂

J


Next segment / lecture …

Band diagram (this segment)


Applying a Bias: Poisson Equation

28

EF-EV

EC-EF

qVbi

EC-Fn

q(Vbi-V)

Fp-EV

-qV

Question: Max value of Vbi?Answer: for degenerate s.c.,

if EC-EF=0, EF-EV=0 � Eg

( )

( )

( )

( )

β

β

−

− −

=

=

i

p i

n E

i

F

i

F

E

n x n e

p x n e

( )2 β−× = pn FF

in p n e

No bias

Appliedbias

P-side grounded


Depletion Widths

29

( )22

0 02 2AD

bi

s

n

s

pqNqN

Vq Vk

xx

k ε ε− = +

D A pnN = N xx ( )

02( )s A

bi

D A D

n

k NV

q N N NVx

ε= −+

( )02

( )s D

bi

A A D

p

k NV

q N N NVx

ε= −+

xn xp

GND-V DN

AN

What about heterojunctions?

From previous lecture (homo-junction)

Applied bias


Fields and Depletion at Forward/Reverse Biases

Charge

Electric Field

Potential

Position

Position

Position

VA<0VA=0

VA>0

VA<0VA=0

VA>0

VA<0VA=0

VA>0

VA<0VA=0

VA>0

NP

Barrier height is reducedat forward biases

Significant increase of peak field at reversebias


Summary PN-Junction Electrostatics

31

1) Learning to draw band-diagrams is one of the most

important topics you learn in this course. Band-diagrams are

a graphical way of quickly solving the Poisson equation.

2) If you consistently follow the rules of drawing band-diagrams,

you will always get correct results. Try to follow the rules, not

guess the final result.


ECE606: Solid State Devicesp-n diode I-V characteristics

Gerhard [email protected]


Outline

33

1) Derivation of the forward bias formula

2) Solution in the nonlinear regime

3) I-V in the ambipolar regime

4) Conclusion

Ref. SDF, Chapter 6


Topic Map

34

Equilibrium DC Small signal

Large Signal

Circuits

Diode

Schottky

BJT/HBT

MOSFET

Diode in Non-Equilibrium(External DC voltage applied)


Continuity Equations for p-n junction Diode

35

1P P P

pr g

t q

∂ = ∇ • − +∂

J

( )+ −∇ • = − + −D AE q p n N N


1N N N

nr g

t q

∂ = ∇ • − +∂

J


This section


Applying a Bias: Poisson Equation

36

EF-EV

EC-EF

qVbi

EC-Fn

q(Vbi-V)

Fp-EV

-qV

Review

No bias

Appliedbias


Depletion Widths

37

( )22

0 02 2AD

bi

s

n

s

pqNqN

Vq Vk

xx

k ε ε− = +

D A pnN = N xx ( )

02( )s A

bi

D A D

n

k NV

q N N NVx

ε= −+

( )02

( )s D

bi

A A D

p

k NV

q N N NVx

ε= −+

xn xp

GND-V DN

AN

Review


Flat Quasi-Fermi Level up to Junction

38

EV

EC

EC

EV

Jn

Jp

Equilibrium: Many electrons N-side. Few electrons P-side.

Detailed balance of drift and diffusion forces. � No net current

Forward Bias: Electron Jn and hole Jp currents across junction.

Diffusion force unchanged. (because doping did not change) But, drift force increased due to applied bias � Net current


Various Regions of I-V Characteristics

39

1. Diffusion limited

2. Ambipolar transport

3. High injection

4. R-G in depletion

5. Breakdown

6. Trap-assisted R-G

7. Esaki Tunneling

VA

2

4

5

1

3

6,7

ln(I)

ln( ) ~ AB

qI V

k T

1

1. Diffusion limited


Recall: One Sided Minority Diffusion

40

1∂ = − +∂

n

n n

dnr g

t q dx

J

µ= +n nn

dnqn qDJ

dxE

2

20 =

n

d nD

dx

Steady stateAcceptor doped

Fp-EV

q(Vbi-V)

-V

If current is continuous, one can calculatefor the current anywhere. Position doesn’tmatter! Calculate at “easiest” position.

�Minority carrier current on P-side.We know the solution from earlierAssume steady state, no rn, gn


Boundary Conditions

41

( )

( )

( 0 )

( 0 )

β

β

−+

− −+

= =

= =

n i

p i

F E

i

F E

i

n x n e

p x n e

( )2 2β β−= =pn AqV

i i

F Fn e n enp

FpFn

q(Vbi-VA)

-VA

2

(0 ) β+ = Aqi

A

Vn e

N

n

(0 )+ =A

p N

AN

( )0

2

(0 ) (0 ) (0 )

1β

+ + +=∆ = −

= −

G G

A

V V

qVi

A

n n n

ne

N

Boundary Conditions

Solution profileDifference of Quasi-Fermi-levels equals applied voltage

X=0+


Right Boundary Condition

42

2

( )

( ) 0

ip

A

p

nn x W

N

n x W

= ≈

∆ = =

EV

EC

Vmetal Infinite surface recombination

velocity at metal � excess carrier concentration = 0


Example: One Sided Minority Diffusion

43

( )2

, ( ) 0

0 ', ( 0) 1A

p p p

qVi

A

x W n x W C DW

nx n x e C

Nβ

= ∆ = = ⇒ = −

= ∆ = = − =

( )2

( , ) 1 1AqVi

A p

n xn x t e

N W

∆ = − −

β

2

20N

d nD

dx=

( , )n x t C Dx∆ = +VAnsatz

Final result: Excess electron carrier concentration (P-side) as function of position

Plug in B.C.


Electron & Hole Fluxes

44

( )2

0

1AqVn in n

x p A

qD ndnJ qD e

dx W Nβ

=

= = − −

( )2

( ) 1 1AqVi

A p

n xn x e

N W

∆ = − −

βN N Nqn qD nµ= + ∇J E

Fp

Fn

∆n

∆p

( )2

0'

1Ap qVip p

x n D

qD ndpJ qD e

dx W Nβ

=

= − = − −

Current (electrons)

Current (holes)


Total Current

45

( )22

1An i

p

qVT

p i

DA n

D n

W N

D n

W NJ q e β

= − + −

ln(I)

VA

(1)(2)

(3)

(4)

(5)

Fp

Fn

∆n

∆p

ln ln( .)T A BJ qV k T const≈ +Forward Bias

.TJ const≈Reverse Bias

To get total current, add electron and hole current values at identical x-points.(assume no recombination)

Diffusion current


Outline

48




4) Conclusion


Nonlinear Regime (3) …

49

( )( )2 2

1n pAqVpn i iT

p A n D

FFDD n nJ q e

W N W N

β−∆ ∆− = − + −

VA

2

1

3

6,7

ln(I)

( )( )0 1A pnJq V a bJI e

β− −= −

Assumption of flat Quasi-Fermi levels invalid here

Today’s lecture: Nonlinear Regime (2,3)


Flat Quasi-Fermi Level up to Junction ?

50

N N N

dnqn qD

dxµ= +J E

) )( (n n ii FF E ni N iN

EdFdnn n e qD qD

dx dxn eβ ββ− − = = −

E

Fp

Fn

n n nn n n

n D

dF J WJ F

dxn

Nµ

µ= ⇒ ∆ =

nN N

n N BN

n

dFdnqD qD

dx dx

dF D k Tq n

dx q

nβ

µµ

= −

= − = ∵

E

E

Wn

Rewrite n into non-equilibrium form, re-arrange Jn equation

New diffusion component: Plug this into original Jn equation

Drop of Quasi-Fermi level across the junction proportional to current!


Forward Bias: Nonlinear Regime …

51

n

p

n

n D

p n

n

n D

J W

N

FN

F

J W

µ

µ∆ =

∆ =

2( )(0 ) n pi

junctio

F

A

F

n

nn e

Nβ−+ =

( )( )2 2

1n pAqVpn i iT

p A n D

FFDD n nJ q e

W N W N

β−∆ ∆− = − + −

∆Fp

∆n

∆Fn

Still diffusion dominated transport? Since Quasi-Fermi levels are not flat in nonlinear regime (drift), this approximation becomes worse.

VA

( ) ( )( )2 2

(0 ) 1pA An pnqV qVi F

A

F Fi

A

Fn ne n e

N N

β β∆ ∆− − − −∆ + ∆= ⇒ ∆ = −


Outline

52




4) Tunneling and I-V characteristics

5) Conclusion


Region (2): Ambipolar Transport

53

( ) / 2ln(

2)A n pqV F Fpn A

T i Tp n B

DD qVJ q n e J

W W k T

β−∆ −∆ ≈ − + ≈

VA

2

1

3

6,7

ln(I)

Today’s lecture: Ambipolar Transport regime (2) Question: Where does the 2 come from?


Nonlinear Regime: Ambipolar Transport

54

( )

( )

/ 2

/ 2

A n p

A n p

qV F Fn in n

p p

qV F Fp ip p

n n

qD nnJ qD e

W W

qD nnJ qD e

W W

β

β

−∆ −∆

−∆ −∆

∆= − =

∆= − =

Note: junction never disappears, even for large forward bias!

( )( )( ) / 2

1A n p

A n p

q V F F

i

q V F F

i

n epn

n e

β

β

−∆ −∆

−∆ −∆≈

∆ = −∆≈

( )( )( )2

22( )( ) 1

n p

A n p

F F

i

q V F FiA i

A

np n e

nN n

Npn e

β

β

−

−∆ −∆∆

=

+ =∆+ −

Here not negligibly small. Ambipolar transport !

Excess carrier concentrations >> NA Thus…

Currents


Outline

55




4) Tunneling and I-V characteristics

5) Conclusion


Forward Bias Nonlinearity (7): Esaki Diode

I

VA

Fn Fp

Fn

Fn

Fp

Fp

X

VA

2

1

3

6,7

ln(I)

Heavy doping

1

empty

No states!

Esaki-Diode: Heavily doped diode

Tunneling in diodes.Nobel Prize (Esaki)


Reverse Bias (5): Zener Tunneling

57

2 2

4

4cosh sinh

(p.49 ADF)

= υ

=α α + − α α

I qpT

Tk

d dk

Fn

Fp

4

5

+VA

empty

empty

Tunneling (triangular barrier)

Remember: Tunneling through a triangular barrier

Zener tunneling occurs in every diode. (reverse bias)


Conclusion

58

1) I-V characteristics of a p-n junction is defined by many

interesting phenomena including diffusion, ambipolar transport,

tunneling etc.

2) The separate regions are identified by specific features. Once

we learn to identify them, we can see if one or the other

mechanism is dominated for a given technology.

3) In the next class, we will discuss a few more non-ideal effects.

ece606: solid state devices lecture 14 …ee606/downloads/ece606_f...klimeck –ece606 fall 2012...

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